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Softa [21]
2 years ago
11

Find the area plzz I don’t remember how to do it

Mathematics
2 answers:
Natasha2012 [34]2 years ago
7 0

Answer:

770,000 is the answer.

Step-by-step explanation:Remember L times W times H or depth. 10 times 5 times 20 than multiply the product by 22, 5, and 7 then that is your answer.

Maru [420]2 years ago
7 0
770,000 is the answer
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Si a es la raíz cuadrada de b ¿también - a será la raíz cuadrada de b?​
Anuta_ua [19.1K]

Answer:

si

Step-by-step explanation:

si

4 0
2 years ago
When a distribution is mound-shaped symmetrical, what is the general relationship among the values of the mean, median, and mode
yuradex [85]

Answer:

The mean, median, and mode are approximately equal.

Step-by-step explanation:

The mean, median, and mode are <em>central tendency measures</em> in a distribution. That is, they are measures that correspond to a value that represents, roughly speaking, "the center" of the data distribution.

In the case of a <em>normal distribution</em>, these measures are located at the same point (i.e., mean = median = mode) and the values for this type of distribution are symmetrically distributed above and below the mean (mean = median = mode).

When a <em>distribution is not symmetrical</em>, we say it is <em>skewed</em>. The skewness is a measure of the <em>asymmetry</em> of the distribution. In this case, <em>the mean, median and mode are not the same</em>, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (<em>negative skew</em>), or greater than them (<em>positive skew</em>), or approximately equal than the median but much greater than the mode (a variation of a <em>positive skew</em> case).  

In the case of the normal distribution, the skewness is 0 (zero).

Therefore, in the case of a <em>mound-shaped symmetrical distribution</em>, it resembles the <em>normal distribution</em> and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, <em>they are all approximately equal</em>. So, <em>the </em><em>general</em><em> relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, </em>considering they have similar characteristics of the <em>normal distribution</em>, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).

5 0
2 years ago
Hey can you please help me posted picture of question
Debora [2.8K]
We can use quadratic formula to determine the roots of the given quadratic equation.

The quadratic formula is:

x= \frac{-b+- \sqrt{ b^{2} -4ac} }{2a}

b = coefficient of x term = -11
a = coefficient of squared term = 2
c = constant term = 15

Using the values, we get:
x= \frac{11+- \sqrt{121-4(2)(15)} }{2(2)} \\  \\ &#10;x= \frac{11+-1}{4} \\  \\ &#10;x=3, x= 2.5

So, the correct answer to this question are option B and D
4 0
3 years ago
How do you compare and order fractions?
AveGali [126]

Answer:

Give each a common denominator

Step-by-step explanation:

So let’s say you have 1/2 and 2/4 how would you know which is bigger well we know both 2 and 4 go into 4 so 1/2 x2 = 2/4 so these 2 are equal

4 0
2 years ago
30 increased by 55% I need help solving this, I am good in history ect, but maths kills me
Lubov Fominskaja [6]
I honestly don't know, but I think it is 63

3 0
3 years ago
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